Post on 15-Jan-2016
Sections
WORKSHEET 9c
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Question 1awhen we want to design a beam
its shape and dimensionsa
its materialb
its Section Modulusc
what do we need to determine?
a and cd
a and be
Question 1bwhen we want to design a beam
the span and support typesa
the total load on the beamb
the shape and dimensions of the beamc
what do we need to know to start?
the maximum bending momentd
the moment of inertia and section moduluse
a and bf
a, b and eg
Question 1cwhen we want to design a beam
the maximum allowable bending stressa
the Modulus of Elasticityb
the maximum allowable deflectionc
what else do we need to know?
whether it is elastic, plastic or brittle d
all the abovee
a, b and cf
Question 2aIn a two-story house, we are spanning across a double garage 7m wide with steel beams at3.6 m centres the floor joists in the upper floor of a house.
The timber joist system in Q 19 spans between these steel beams (200 x 50 mm softwood
joists @ 600 mm centres spanning 3.6 m.) The live load is 1.5 kPa and the dead load is 0.4
kPa (including the self-weight of the joist)
what is the tributary area?
4.2 m2a
49.0 m2b
25.2 m2c
Question 2bIn a two-story house, we are spanning across a double garage 7m wide with steel beams at3.6 m centres the floor joists in the upper floor of a house.
The timber joist system in the previous tutorial spans between these steel beams
(200 x 50 mm softwood joists @ 600 mm centres spanning 3.6 m.) The live load is 1.5 kPa and
the dead load is 0.4 kPa (including the self-weight of the joist)
what is the total load on a beam?
47.9 kNa
100.8 kNb
47.9 kPac
Question 2c
is this load a?
uniformly distributed load (UDL)a
a point loadb
In a two-story house, we are spanning across a double garage 7m wide with steel beams at3.6 m centres the floor joists in the upper floor of a house.
The timber joist system in the previous tutorial spans between these steel beams
(200 x 50 mm softwood joists @ 600 mm centres spanning 3.6 m.) The live load is 1.5 kPa and
the dead load is 0.4 kPa (including the self-weight of the joist)
Question 2d
do we design for strength or for stiffness
stiffnessa
strengthb
In a two-story house, we are spanning across a double garage 7m wide with steel beams at3.6 m centres the floor joists in the upper floor of a house.
The timber joist system in the previous tutorial spans between these steel beams
(200 x 50 mm softwood joists @ 600 mm centres spanning 3.6 m.) The live load is 1.5 kPa and
the dead load is 0.4 kPa (including the self-weight of the joist)
Question 2e
in order to get the dimensions of the beamwhat do we need to find?
the stress in the beama
the minimum required Section Modulusb
the minimum required Moment of Inertiac
In a two-story house, we are spanning across a double garage 7m wide with steel beams at3.6 m centres the floor joists in the upper floor of a house.
The timber joist system in the previous tutorial spans between these steel beams
(200 x 50 mm softwood joists @ 600 mm centres spanning 3.6 m.) The live load is 1.5 kPa and
the dead load is 0.4 kPa (including the self-weight of the joist)
Question 2f
in order to find the minimum requiredSection Modulus, what do we need to find?
the maximum Bending Momenta
the stress in the beamb
the maximum Shear Forcec
In a two-story house, we are spanning across a double garage 7m wide with steel beams at3.6 m centres the floor joists in the upper floor of a house.
The timber joist system in the previous tutorial spans between these steel beams
(200 x 50 mm softwood joists @ 600 mm centres spanning 3.6 m.) The live load is 1.5 kPa and
the dead load is 0.4 kPa (including the self-weight of the joist)
Question 2g
what is the maximum Bending Moment?
293.3 kNma
83.8 kNmb
41.9 kNmc
In a two-story house, we are spanning across a double garage 7m wide with steel beams at3.6 m centres the floor joists in the upper floor of a house.
The timber joist system in the previous tutorial spans between these steel beams
(200 x 50 mm softwood joists @ 600 mm centres spanning 3.6 m.) The live load is 1.5 kPa and
the dead load is 0.4 kPa (including the self-weight of the joist)
In a two-story house, we are spanning across a double garage 7m wide with steel beams at3.6 m centres the floor joists in the upper floor of a house.
The timber joist system in the previous tutorial spans between these steel beams
(200 x 50 mm softwood joists @ 600 mm centres spanning 3.6 m.) The live load is 1.5 kPa and
the dead load is 0.4 kPa (including the self-weight of the joist)
Question 2h
we want to find a Universal Beam which is strong enough. The maximum allowable stress of grade 300 steel is 200 MPa
what is the minimum required Section Modulus?
209.5 x 103 mm3a
477.3 x 103 mm3b
4.8 x 103 mm3c
Question 2iwe want to find a Universal Beam which is strong enough. Given the we need a Section
Modulus of at least 209.5 x 103 mm3, we can look up a Table of Universal Beams.
which beam would we select as being strong enough?
200 UB 29.8a
200 UB 25.4b
180 UB 22.2c
click here to see the Table of Universal Beams
250 UB 31.4d
back toQ2i
back toQ2n
back toQ2t
Question 2jhaving found a beam section 200UB 25.4 (with a depth of 203 mm) as being strong enough
what do we need to do now?
use ita
check it for strengthb
check it for stiffnessc
Question 2kto check the stiffness of a beam
what must we do?
check its widtha
check its deflectionb
check its span-to-depth ratioc
Question 2lto check the deflection of a beam
what must we do?
compare the actual deflection with the maximum allowable deflection
a
check that it doesn’t deflect more than thespan-to-depth ratio
b
make the beam twice as deep c
Question 2mto check the deflection of the 200 UB 25.4 beam
what else do we need to know?
the total load and the span a
the Moment of Inertia (I) and the Modulus of Elasticity, E
b
the maximum allowable deflectionc
b and cd
a, b and ce
Question 2ngiven that the Modulus of Elasticity of structural steel is 200000 MPa and the maximum allowable deflection is span / 500, we need to find the Moment of Inertia of the200 UB 25.4 beam.
We can do that by going back to the Table of Properties
what is the Moment of Inertia of the beam?
28.9 x 106 mm4 a
44.4 x 106 mm4 b
23.6 x 106 mm4c
click here to see the Table of Universal Beams
Question 2oGiven that the Moment of Inertia of the beam is 23.6 x 106 mm4 and that the Modulus of Elasticity of structural steel is 200000 MPa, we need to find the deflectionThe deflection formula for a simply supported beam with a uniformly distributed load is
5 x WL3 / 384 E I
(remember the total load is 47.88 kN, the span is 7 m)
what is the deflection of the beam?
45.3 mma
453.0 mm b
35.5 mmc
Question 2pGiven that the deflection of the beam is 45.3 mm and the maximum allowable deflection is span / 500
is the section stiff enough?
yesa
no b
Question 2qso the section is not stiff enough
what do we need to do?
upsize it by increasing its Section Modulusa
upsize it by increasing its Moment of Inertia b
increase its Modulus of Elasticityc
Question 2rso the section is not stiff enough.
A Moment of Inertia of 23.6 x 106 mm4 produced a deflection of 45.3 mm The maximum allowable deflection is 14.0 mm
how much stiffer must the section be ?By how much must we increase
the Moment of Inertia?
by a factor of 3.24a
by a factor of 1.92 b
by a factor of 1.69c
Question 2sso the section is not stiff enough.
A Moment of Inertia of 23.6 x 106 mm4 produced a deflection of 45.3 mm
So we need to increase the Moment of Inertia by a factor of 3.24
what value of the Moment of Inertia do we need?
330.4 x 106 mm4a
45.3 x 106 mm4b
76.4 x 106 mm4c
Question 2tSo we need a section that has a Moment of Inertia of 76.4 x 106 mm4
We need to go back to the Table of Properties to find a satisfactory section
what section has a satisfactory Moment of Inertiaand, therefore is stiff enough?
250UB 37.3a
310UB 46.2b
310UB 40.4c
click here to see the Table of Universal Beams
next question
enough !
when we design anything what we are doing is determining its material, shape and dimensions
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that’s only part of it
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we will need to use the Section Modulusbut that’s not what we are aiming at
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enough !
yes, we always need to know what the span, support type and loads are
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that’s only part of it
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isn’t that what we are trying to find?
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we will need to calculate the maximum Bending Momentbut we can do that from the information we need to know
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we will need to find the minimum required I and Zbut we can calculate that from the information we will have
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enough !
in order to calculate some of the things that we will have towe need to be given the maximum allowable bending stress,
the Modulus of elasticity and the maximum allowable deflection.This we will get from codes once we decide on a material
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that’s only part of it
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Why would we want to know whether the beam will be ofan elastic, plastic or brittle material at this stage?
What would that give us?
next question
enough !
3.6 m7
m
steel beams
timber joists@ 600mm crs
3.6 m
tributary areafor beam = 7 x 3.6 = 25.2 m2
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How did you get that? The beams span 7m and they are 3.6m centres
The load on the beam (neglecting the weight of the joists) isthe sum of the live load (1.5kPa) and the dead load (0.4kPa)
next question
enough !
Tributary area = 25.2 m2
load per sq m = 1.9 kPa
TOTAL LOAD = 25.2 x 1.9 = 47.9 kN (kPa x m2 = kN)
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How did you get that?What is the total load per sq m?
What is the tributary area?
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you should learnWe are talking about total load.
So what are the units for a load (force)?
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enough !
that’s rightthe load is distributed over the length of the beam
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is the load acting at just one pointor is it acting all along the length of the beam?
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enough !
we design for strength and check for stiffness
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No, not really
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enough !
once we have the Z value we can find the dimensionssince we are designing for strength we need to find the
minimum required Section Modulus since that is the propertyof a beam’s section which determines its strength
i.e. a beam with the minimum required Z will be strong enough.
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we already have that given as the maximum allowable bending stress
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The Moment of Inertia of a beam’s X-section determines its stiffnessNOT its strength
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enough !
the formula for bending stress is f = M / Z (stress = BM / Section Modulus)
so Z = M / fsince we know f (max allowable bending stress)
we need to first find the maximum BM
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we already have that given as the maximum allowable bending stress
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what has the shear force to do withthe Section Modulus?
next question
enough !
since this is a simply supported beam with a UDL Max BM = W L / 8 (where W = total load)
= 47.88 x 7 / 8 = 41.9 kNm
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How did you get that? Don’t guess!what is the max BM formula for a simply supported beam with a UDL?
What is the total load? (see Q2b) What is the span?
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What is the max BM formula for asimply supported beam with a UDL?
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enough !
Z = M / f = 41.9 / 200 (since M is in kNm bring to N and mm)
= 41.9 x 106 / 2 x 102
= 209.5 x 103 mm3
How did you get that? What did we say the formula for Z was?
What is the max BM?
What is the stress that we can allow the beam to take?
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next question
enough !
a universal beam 200UB 25.4 has a Zx value of 232 x 103 mm3
since this > the min required Zx of 209.5 x 103 mm3 it is strong enough.A smaller beam would not be strong enough and
a larger beam would be wasteful
The Zx value is certainly adequate andthe beam would be strong enough
But it is more than necessary and would be wasteful
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The Zx value of a 180 UB 22.2 is only 171 x 103 mm3
such a beam would not be strong enough
since we need at least a Zx value of 209.5 x 103 mm3
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next question
enough !
Yes, the beam may be strong enoughbut maybe it deflects too much
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It is strong enough but what else could go wrong?
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but we’ve already designed it for strengthwe know that it’s strong enough
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enough !
Exactly! We have to compare its actual deflection withthe maximum allowable deflection
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what will that give us?
We could do that. That will give us an indicationBut we really need to be precise.
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next question
enough !
this will tell us if the beam is stiff enough
how do you compare a deflection with a span-to depth ratio?One is a distance, the other is a ratio
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how do you know whether that’s enough or too much?
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next question
enough !
remembering the deflection formula d = k x WL3 / E Iwe see that we need to also know E and I
and of course we need to know the max allowable deflection
we already know the total load and the span
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It’s not the whole story
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next question
enough !
from the Table we see that a 200UB 25.4
has a value of Ix of 23.6 x 106 mm4
check againIt’s a 200 UB 25.4 we are looking at
not a 200 UB 29.8
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check againIt’s a 200 UB 25.4 we are looking at
not a 250 UB 31.4
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next question
enough !
= 5 x W L3 / (384 x E I)
= 5 x 47.88 x 73 / (384 x 2 x 105 x 23.6 x 106)(the load is in kN and the span is in m, so bring everything to N & mm)
= 5 x 47.88 x 103 x (7 x 103)3 / (384 x 2 x 105 x 23.6 x 106)
= 5 x 47.88 x 343 x 1012 / (384 x 2 x 23.6 x 1011) = 82114.2.84 x 10 / 18124.8 = 45.3 mm
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don’t you think that that’s a rather large deflection
check your zeros
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work it out
you have all the necessary values
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enough !
the actual deflection is 45.3 mmthe max allowable deflection is span / 500 = 7000 / 500 = 14 mm
so the actual deflection is much greater than the max allowable deflectionso the section is not stiff enough
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what is the deflection of this beam?what is the maximum allowable deflection?
is the actual deflection less or greater than the max allowable deflection?
next question
enough !
that’s exactly it.increasing the Moment of Inertia of a beam’s section
will increase its stiffness
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we are talking about stiffness – not strength
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theoretically we could but where would we find a materialwith such a high Modulus of Elasticity?
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enough !
the actual deflection is 45.3 mmthe deflection we want is 14.0 mmso we want the beam to be 45.3 / 14 times stifferso we must increase the Moment of Inertia by a factor of 3.24
work it out
what is the actual deflection?what is the maximum allowable deflection that we want?
so how much stiffer must the beam be?
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next question
enough !
the Moment of Inertia of the 200UB25.4 is 23.6 x 106 mm4
we need to increase that by a factor of 3.24
this gives us a required I of 23.6 x 3.24 = 76.4 x 106 mm4
work it out
what is the Moment of Inertia of the 200UB 25.4?what is the factor by which we have to increase it?
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You’ve graduated with honours!
FINISH
a 310UB 40.4 has an Ix value of 85.2 x 106 mm4 which is the lowest we can find that is
greater than the required min value of 76.4 x 106 mm4
look again
the 250UB 37.3 has an Ix of 55.6 x 106 mm4
but we need an Ix of at least 76.4 x 106 mm4
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look again
the 310UB 46.2 has an Ix of 99.5 x 106 mm4
that is greater than 76.4 x 106 mm4 but somewhat wasteful as we could use a smaller beam.
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